Minimal Ring Generated by System of Sets

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Theorem

Let $\mathcal S$ be a non-empty system of sets.


Then there is a unique ring of sets $\mathcal R \left({\mathcal S}\right)$ which:

$(1): \quad$ contains $\mathcal S$
$(2): \quad$ is contained by every ring of sets which also contains $\mathcal S$.


This ring of sets $\mathcal R \left({\mathcal S}\right)$ is called the minimal ring generated by $\mathcal S$.


Proof

Uniqueness

Suppose there were two such rings of sets $\mathcal R \left({\mathcal S}\right)$ and $\mathcal R \left({\mathcal S}\right)'$.

Then by definition their intersection $\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)'$ would also contain $\mathcal S$.

By Intersection of Rings of Sets, $\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)'$ is also a ring of sets.

From Intersection is Subset, $\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)' \subseteq \mathcal R \left({\mathcal S}\right)$ and $\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)' \subseteq \mathcal R \left({\mathcal S}\right)'$.

Hence either $\mathcal R \left({\mathcal S}\right)$ or $\mathcal R \left({\mathcal S}\right)'$ can not be minimal.

So if $\mathcal R \left({\mathcal S}\right)$ exists, it has to be unique.


Existence

Consider the union:

$\displaystyle X = \bigcap_{A \in \mathcal S} A$

of all sets in $\mathcal S$.

Now consider the power set $\mathcal P \left({X}\right)$ of all subsets of $X$.

From Power Set is Algebra of Sets and by definition of algebra of sets, $\mathcal P \left({X}\right)$ is a ring of sets containing $\mathcal S$.

Now let $\Sigma$ be the set of all rings of sets contained in $\mathcal P \left({X}\right)$ which also contain $\mathcal S$.

Then consider the intersection:

$\displaystyle \mathcal R \left({\mathcal S}\right) = \bigcap_{\mathcal T \in \Sigma} \mathcal T$

of all these rings of sets.

It is clear that $\mathcal R \left({\mathcal S}\right)$ has the properties specified.

In particular, as all $\mathcal T \in \Sigma$ contain $\mathcal S$, then so does $\mathcal R \left({\mathcal S}\right)$.


Finally, note that if $\mathcal R'$ is a ring of sets containing $\mathcal S$, then:

$\mathcal R^* = \mathcal R' \cap \mathcal P \left({X}\right)$

is a ring of sets in $\Sigma$.

Hence $\mathcal R \left({\mathcal S}\right) \subseteq \mathcal R^* \subseteq \mathcal R'$ as we needed to show.

$\blacksquare$